This page 10th samacheer kalvi math solution for exercise 5.3 part 2 is going to provide you solution for every problems that you find in the exercise no 5.3

4. Find the angle of inclination of the line passing through the points

(i) (1, 2) and (2 , 3)

Solution:

**x****₁ = 1 ** ** ****x****₂ = 2**

**y₁ = 2 ** ** ****y₂ = 3**

**(y-y**** ₁**)/(y

** (y - 2)/(3 - 2) = (x - 1)/(2 - 1)**

** (y - 2)/1 = (x - 1)/1**

** y - 2 = x - 1**

** x - y - 1 + 2 = 0 **

** x - y + 1 = 0**

(ii) (3 , 3) and (0 , 0)

x₁ = 3 x₂ = 0

y₁ = 3 y₂ = 0

(y-y₁)/(y₂-y₁) = (x-x₁)/(x₂-x₁)

(y - 3)/(0 - 3) = (x - 3)/(0 - 3)

(y - 3)/(-3) = (x - 3)/(-3)

- 3 (y - 3) = -3 (x - 3)

- 3 y + 9 = - 3 x + 9

3 x - 3 y + 9 - 9 = 0

3 x - 3 y = 0

divide the whole equation by 3

x - y = 0

(iii) (a , b) and (-a , -b)

x₁ = a x₂ = -a

y₁ = b y₂ = -b

(y-y₁)/(y₂-y₁) = (x-x₁)/(x₂-x₁)

(y - b)/(-b - b) = (x - a)/(-a - a)

(y - b)/(-2b) = (x - a)/(-2a)

- 2a (y - b) = -2b (x - a)

-a y + ab = -b x + ab

b x - a y + ab - ab = 0

b x - a y = 0

5. Find the slope of the line which passes through the origin and the midpoint of the line segment joining the points (0 ,- 4) and (8 , 0).

Solution:

midpoint of the line segment joining the points (0,-4) and (8 , 0)

midpoint = (x₁ + x₂)/2 , (y₁ + y₂)/2

x₁ = 0 x₂ = 8

y₁ = -4 y₂ = 0

= (0 + 8)/2 , (-4 + 0)/2

= 8/2 , -4/2

= (4 , -2)

now we need to find the slope of the line which passes through origin

x₁ = 4 x₂ = 0

y₁ = -2 y₂ = 0

m = (y₂ - y₁)/(x₂ - x₁)

= (0 - (-2))/(0 - 4)

= 2/(-4)

= -1/2

In the page 10th samacheer kalvi math solution for exercise 5.3 part 2 we are going to see the solution of next problem

6. The side AB of a square ABCD is parallel to x-axis . Find the

(i) slope of AB (ii) slope of BC (iii) slope of the diagonal AC

Solution:

(i) Since the side AB is parallel to x axis,

slope of the side AB = 0

(ii) The angle formed by the side BC is 90.

m = tan θ

θ = 90

m = tan 90

= ∞

(iii) The diagonal AC is the angle bisector of the angle.

m = tan θ

θ = 45

m = tan 45

= 1

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